Problem: Solve for $x$ and $y$ using substitution. ${4x-3y = -12}$ ${y = 6x-10}$
Since $y$ has already been solved for, substitute $6x-10$ for $y$ in the first equation. ${4x - 3}{(6x-10)}{= -12}$ Simplify and solve for $x$ $4x-18x + 30 = -12$ $-14x+30 = -12$ $-14x+30{-30} = -12{-30}$ $-14x = -42$ $\dfrac{-14x}{{-14}} = \dfrac{-42}{{-14}}$ ${x = 3}$ Now that you know ${x = 3}$ , plug it back into $\thinspace {y = 6x-10}\thinspace$ to find $y$ ${y = 6}{(3)}{ - 10}$ $y = 18 - 10$ $y = 8$ You can also plug ${x = 3}$ into $\thinspace {4x-3y = -12}\thinspace$ and get the same answer for $y$ : ${4}{(3)}{ - 3y = -12}$ ${y = 8}$